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8 months ago

Anna Litical analyzes the force between a planet and its moon, varying the mass of

Physics

1 answer:

Travka [436]8 months ago

5 0

Answer:

Trial 1 is the largest, trial 3 is the smallest

Explanation:

Given:

Trial 1

M₁ = 6·10²² kg

d₁ = 3 500 km = 3.5·10⁶ м

Trial 2

M₂ = 6·10²² kg

d₂ = 7 000 km = 7·10⁶ м

Trial 3

M₃ = 3·10²² kg

d₃ = 7 000 km = 7·10⁶ м

___________

F - ?

Gravitational force:

F₁ = G·m·M₁ / d₁² = m·6.67·10⁻¹¹·6·10²² / (3.5·10⁶)² = 0.37·m (N)

F₂ = G·m·M₂ / d₂² = m·6.67·10⁻¹¹·6·10²² / (7·10⁶)² = 0.08·m (N)

F₃ = G·m·M₃ / d₃² = m·6.67·10⁻¹¹·3·10²² / (7·10⁶)² = 0.04·m (N)

Trial 1 is the largest, trial 3 is the smallest

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To increase the potential energy of any object, not limited to an apple, you have to hold it higher.

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3 years ago

Define thermopile and peltiers??

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Answer:

Thermopile is an electronic device that converts thermal energy into electrical energy.

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2 years ago

If 190 grams of water is cooled from 42.7°C to 21.2° C how much energy was lost by the water?

yanalaym [24]

Answer:

Energy lost by the water is 17.1 x 10³ J .

Explanation:

Specific heat is defined as the amount of energy per unit mass needed to raise the temperature of the substance by a one degree Celsius.

Heat energy gain or loss by any substance is given by :

Q = m x C x ( T₁ - T₂ ) .....(1)

Here m is the mass of the substance, C is specific heat of the substance and T₁ and T₂ are the initial and final temperature of the substance.

According to the problem,

Mass of water, m = 190 gm

Specific heat of water, C = 4.186 J gm⁻¹ ⁰C⁻¹

Initial temperature, T₁ = 42.7⁰ C

Final temperature, T₂ = 21.2⁰ C

Substitute these values in equation (1).

Heat energy loss by water = 190 x 4.186 x ( 21.2 - 42.7 )

= -17.1 x 10³ J

In the above value, negative sign denotes the loss in energy.

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3 years ago

How does the shape and size of the continents today help support the Theory of Continental Drift?

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3 years ago

Suppose the coefficient of static friction between a quarter and the back wall of a rocket car is 0.383. At what minimum rate wo

djverab [1.8K]

Answer:

25.59 m/s²

Explanation:

Using the formula for the force of static friction:

$f_s = \mu_s N$ --- (1)

where;

$f_s =$ static friction force

$\mu_s =$ coefficient of static friction

N = normal force

Also, recall that:

F = mass × acceleration

Similarly, N = mg

here, due to min. acceleration of the car;

$N = ma_{min}$

From equation (1)

$f_s = \mu_s ma_{min}$

However, there is a need to balance the frictional force by using the force due to the car's acceleration between the quarter and the wall of the rocket.

Thus,

$F = f_s$